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Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− , and the given pKa value for propanoic acid in the expression. pH= 4.874 + Log [A-]/[HA] Using the equation to calculate the quotient [A−] / [HA] at three different pH values.(a) pH = 4.23(b) pH = 4.87(c) pH = 5.3

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Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

  • CH₃CH₂CO₂H = HA
  • CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

  • (a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

[tex]10^{-0.644}[/tex] = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

  • (b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

[tex]10^{-0.004}[/tex] = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

  • (c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

[tex]10^{0.426}[/tex] = [A⁻]/[HA]

2.667 = [A⁻]/[HA]

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